The Beta Density and Beta Function

Why the Beta Function Appears

A probability density must integrate to 1 over its support. For a function proportional to $x^{a-1}(1-x)^{b-1}$ on the unit interval, the integral depends on both positive parameters $a$ and $b$. The beta function names this integral, so it gives the normalizing constant directly. The beta–gamma identity then provides a way to evaluate it using the gamma function.

This tutorial develops that normalization carefully. The examples focus on identifying the parameters, finding the integral, and checking the resulting density. They do not require computing moments or probabilities over subintervals.

Definition: For parameters $a>0$ and $b>0$, the beta function is $$ B(a,b)=\int_0^1 x^{a-1}(1-x)^{b-1}\,dx. $$ A random variable $X$ has a beta distribution with parameters $a,b$, written $X\sim\operatorname{Beta}(a,b)$, if its density is $$ f_X(x)=\frac{x^{a-1}(1-x)^{b-1}}{B(a,b)},\qquad 0<x<1, $$ and $f_X(x)=0$ outside that interval.

The requirements $a>0$ and $b>0$ ensure that the integral defining $B(a,b)$ is finite. The density is positive on the open interval $(0,1)$, and its integral there is 1 because its numerator integrates to $B(a,b)$. Values assigned to a density at the individual endpoints do not affect probabilities; the displayed density formula is stated on the open interval.

More generally, if a nonnegative kernel $g(x)=x^{a-1}(1-x)^{b-1}$ is given on $(0,1)$, a density proportional to it must have the form $c\,g(x)$. Its integral equals 1 precisely when $c=1/B(a,b)$. A proposed coefficient is not automatically a normalizing constant: it must be checked against the integral.

The Gamma Function and the Beta–Gamma Identity

The gamma function extends the factorial to positive real arguments. For $s>0$, define

\Gamma(s)=\int_0^\infty t^{s-1}e^{-t}\,dt.

Integration by parts gives $\Gamma(s+1)=s\Gamma(s)$. In particular, for a positive integer $n$, $\Gamma(n)=(n-1)!$. The standard value $\Gamma(1/2)=\sqrt{\pi}$ follows from evaluating the Gaussian integral. These facts allow many beta-function values to be reduced to factorials and powers of $\pi$.

Theorem: For $a>0$ and $b>0$, the beta and gamma functions satisfy $$ B(a,b)=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}. $$ Consequently, the beta density’s normalizing constant is $$ \frac{1}{B(a,b)}=\frac{\Gamma(a+b)}{\Gamma(a)\Gamma(b)}. $$

The identity follows by writing the product of two gamma integrals as a double integral and changing variables. This derivation also explains why the sum $a+b$ appears in the denominator.

Multiply the gamma integrals.
Because both integrands are nonnegative, the product can be written as a double integral over $u>0$ and $v>0$: $$ \Gamma(a)\Gamma(b)=\int_0^\infty\int_0^\infty u^{a-1}v^{b-1}e^{-(u+v)}\,du\,dv. $$

Separate the total size from the proportion.
Set $r=u+v$ and $x=u/(u+v)$. Then $u=rx$, $v=r(1-x)$, with $r>0$ and $0<x<1$. The absolute Jacobian determinant is $r$, so $du\,dv=r\,dr\,dx$.

Factor the transformed integral.
Substitution gives $$ \Gamma(a)\Gamma(b) =\int_0^\infty\int_0^1 r^{a+b-1}e^{-r}x^{a-1}(1-x)^{b-1}\,dx\,dr =\Gamma(a+b)B(a,b). $$

Rearrange.
Since $\Gamma(a+b)>0$, divide both sides by it to obtain $B(a,b)=\Gamma(a)\Gamma(b)/\Gamma(a+b)$, as claimed.

The change of variables covers the positive quadrant once: $r$ records the sum $u+v$, while $x$ records the fraction of that sum contributed by $u$. The Jacobian factor $r$ is essential. Omitting it would change the power of $r$ and would not produce $\Gamma(a+b)$.

Worked Applications of Normalization

Worked Example: Normalizing an Integer-Parameter Kernel

In a hypothetical reliability calculation, a model assigns a quantity $X$ to the interval $(0,1)$ with density proportional to $x(1-x)^2$. Find the normalized density and evaluate it at $x=1/2$. State the distribution and parameters.

Method. Match the powers to $x^{a-1}(1-x)^{b-1}$. Here $a-1=1$, so $a=2$, and $b-1=2$, so $b=3$. Both parameters are positive, so the beta integral is finite.

Using the beta–gamma identity and $\Gamma(n)=(n-1)!$, $$ B(2,3)=\frac{\Gamma(2)\Gamma(3)}{\Gamma(5)} =\frac{1!\,2!}{4!} =\frac{1\cdot2}{24} =\frac{1}{12}. $$

The required coefficient is the reciprocal, $1/B(2,3)=12$. Thus $X\sim\operatorname{Beta}(2,3)$ and $$ f_X(x)=12x(1-x)^2,\qquad 0<x<1. $$ At $x=1/2$, $$ f_X(1/2)=12\left(\frac12\right)\left(1-\frac12\right)^2 =12\cdot\frac12\cdot\frac14 =\frac32=1.5. $$

Check. Direct integration gives $$ \int_0^1 12x(1-x)^2\,dx =12\left(\frac12-\frac23+\frac14\right) =12\left(\frac{6-8+3}{12}\right)=1. $$ The density is normalized. A common mistake is to use $B(2,3)=1/12$ itself as the coefficient; that would make the integral $1/144$, not 1.

Worked Example: A Half-Integer Parameter

A hypothetical model uses a density proportional to $x^{-1/2}(1-x)^{1/2}$ on $(0,1)$. Find its normalizing constant, identify the beta distribution, and evaluate the density at $x=1/4$. Give the density value to four decimal places.

Method. Comparing exponents gives $a-1=-1/2$, hence $a=1/2$, and $b-1=1/2$, hence $b=3/2$. Both are positive. Apply the beta–gamma identity: $$ B\left(\frac12,\frac32\right) =\frac{\Gamma(1/2)\Gamma(3/2)}{\Gamma(2)}. $$

By the recurrence relation, $\Gamma(3/2)=(1/2)\Gamma(1/2)=\sqrt{\pi}/2$, while $\Gamma(2)=1$. Therefore $$ B\left(\frac12,\frac32\right) =\frac{\sqrt{\pi}(\sqrt{\pi}/2)}{1} =\frac{\pi}{2}. $$ The normalizing constant is $2/\pi$, and $X\sim\operatorname{Beta}(1/2,3/2)$.

At $x=1/4$, the unnormalized kernel is $$ \left(\frac14\right)^{-1/2}\left(\frac34\right)^{1/2} =2\cdot\frac{\sqrt{3}}{2} =\sqrt{3}. $$ Thus $$ f_X(1/4)=\frac{2\sqrt{3}}{\pi}\approx 1.1027. $$ The approximation is rounded to four decimal places.

Check. By the definition of $B$, the integral of the kernel is $\pi/2$; multiplying by $2/\pi$ gives 1. The main trap is treating $\Gamma(3/2)$ as $\Gamma(1/2)$, rather than applying the recurrence factor $1/2$.

Worked Example: Correcting a Proposed Coefficient

For a hypothetical bounded score $X$, an analyst proposes $5\sqrt{x}(1-x)$ as a density on $(0,1)$. Determine whether it integrates to 1. If not, find the correct density and evaluate it at $x=4/9$.

Method. Write the powers as $x^{1/2}(1-x)^1$. This gives $a-1=1/2$, so $a=3/2$, and $b-1=1$, so $b=2$. The integral of the kernel is $$ B\left(\frac32,2\right) =\frac{\Gamma(3/2)\Gamma(2)}{\Gamma(7/2)}. $$

The needed gamma values are $$ \Gamma(3/2)=\frac{\sqrt{\pi}}{2},\qquad \Gamma(2)=1,\qquad \Gamma(7/2)=\frac52\cdot\frac32\cdot\frac12\sqrt{\pi} =\frac{15\sqrt{\pi}}{8}. $$ Therefore $$ B\left(\frac32,2\right) =\frac{\sqrt{\pi}/2}{15\sqrt{\pi}/8} =\frac{4}{15}. $$

The proposed function integrates to $5(4/15)=4/3$, so it is not a density. The coefficient that makes the integral 1 is $1/B(3/2,2)=15/4$. The corrected distribution is $X\sim\operatorname{Beta}(3/2,2)$, with $$ f_X(x)=\frac{15}{4}\sqrt{x}(1-x),\qquad 0<x<1. $$ At $x=4/9$, $\sqrt{x}=2/3$ and $1-x=5/9$, so $$ f_X(4/9)=\frac{15}{4}\cdot\frac23\cdot\frac59 =\frac{150}{108} =\frac{25}{18}. $$

Check. The corrected coefficient times the kernel integral is $(15/4)(4/15)=1$. The mistake to avoid is assuming a plausible-looking coefficient is a normalizer without integrating the function.

Practical Checks and Common Traps

For a proposed beta density, the following sequence keeps the parameter matching and normalization separate:

Confirm that the support is $(0,1)$ and that the kernel has the form $x^{a-1}(1-x)^{b-1}$.
Match each exponent carefully: the parameter is one more than its corresponding exponent.
Check $a>0$ and $b>0$, then evaluate $B(a,b)$.
Multiply the kernel by $1/B(a,b)$, not by $B(a,b)$, and verify that the integral is 1.

An exponent of $-1/2$, for example, corresponds to a parameter of $1/2$, not $-1/2$. Another frequent error is to forget that the beta function integrates over the unit interval; a kernel on a different interval is not put into this form without first accounting for the transformation of the variable and its density.

The beta function is useful because it turns a normalization problem into a named integral, while the beta–gamma identity converts that integral into gamma values. In actuarial and statistical models, this gives a systematic way to check a proposed continuous distribution on a bounded proportion or score scale. For a distribution whose variable is supported on the unit interval, normalization is the first required check; later calculations depend on having a valid density.

Key takeaway: For $a,b>0$, the kernel $x^{a-1}(1-x)^{b-1}$ on $(0,1)$ has integral $B(a,b)$. Divide by that integral to obtain a beta density, and evaluate it efficiently using $B(a,b)=\Gamma(a)\Gamma(b)/\Gamma(a+b)$.

Check Your Understanding

For each problem, identify the beta parameters where applicable and check the normalization rather than assuming the given coefficient is correct.

A hypothetical score has kernel $x^2(1-x)$ on $(0,1)$. Find the normalizing constant and the density value at $x=1/2$. (Answer: $a=3,\ b=2$; constant $12$; density value $3/2$.)
Find the normalizing constant for the kernel $x^{-1/2}(1-x)^{-1/2}$ on $(0,1)$, using the beta–gamma identity. (Answer: $1/\pi$; the parameters are $a=b=1/2$.)
A proposed density is $c\,x(1-x)^{-1/2}$ on $(0,1)$. Find $c$ and identify its beta distribution. (Answer: $a=2,\ b=1/2;\ B(2,1/2)=4/3;\ c=3/4$; $X\sim\operatorname{Beta}(2,1/2)$.)
Normalize the kernel $x^{1/2}(1-x)^{3/2}$ on $(0,1)$. (Answer: $a=3/2,\ b=5/2;\ B(3/2,5/2)=\pi/16$; constant $16/\pi$.)

What You'll Learn